Proof of Gauss's Law


by Swapnil
Tags: gauss, proof
Galileo
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#19
Sep3-06, 10:00 AM
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Quote Quote by leright
BY DEFINITION, a radial field HAS NO DIVERGENCE except at r = 0.
Not true. Radial means that [itex]\vec F(\vec r)=F(r)\hat r[/itex], so the field just points away radially from the origin (or wherever the source may be).
A field like [itex]\vec F(\vec r)=K \hat r/r^3[/itex] is radial, but not divergenceless. That is a property unique to the inverse square relationship. In fact, an inverse square field is the only radial field that is divergenceless.
leright
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Sep3-06, 10:40 AM
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Quote Quote by Galileo
Not true. Radial means that [itex]\vec F(\vec r)=F(r)\hat r[/itex], so the field just points away radially from the origin (or wherever the source may be).
A field like [itex]\vec F(\vec r)=K \hat r/r^3[/itex] is radial, but not divergenceless. That is a property unique to the inverse square relationship. In fact, an inverse square field is the only radial field that is divergenceless.
ok, this is true. There can be radial fields with divergence of zero everywhere, if for instance, the magnitude of the field at r = 0 is 0. I had my mind in electrostatics world and was visualizing the electric fields due to charges, and not just fields in general.
Swapnil
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Sep3-06, 04:53 PM
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Quote Quote by leright
It seems ...I suppose.
Thank you leright for your explanation. I think I have a more better understanding of divergence now.

Quote Quote by Galileo
You could do it in cartesian coordinates:
[tex]\vec F(x,y,z)=K \frac{\hat r}{r^2}=K \frac{\vec r}{r^3}=K \frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2+y^2+z^2)^{3/2}}[/tex]
but that's really a bit of a hassle.
Try it in spherical coordinates, whhich makes it trivial.
I think I am wrong, but isn't [tex]\vec{r}[/tex] the follwoing in spherical coordinates?:

[tex] \vec{r} = r\,\hat{r} + \arctan(\frac{y}{x})\,\hat{\phi} + \arcsin(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}})\,\hat{\theta}[/tex]

How is this more easier?
Galileo
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Sep3-06, 11:13 PM
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No way. In spherical coordinates the field is simply [itex]\hat r/r^2[/itex]. There is no phi or theta dependence in there, since it's radial remember? All you have to do is lookup the divergence in spherical coordinates and do a simply derivative.
Swapnil
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Sep4-06, 12:43 AM
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Quote Quote by Galileo
No way. In spherical coordinates the field is simply [itex]\hat r/r^2[/itex]. There is no phi or theta dependence in there, since it's radial remember? All you have to do is lookup the divergence in spherical coordinates and do a simply derivative.
Oh yeah! Sorry, I guess I wasn't thinking straight now that I think about it...
jbusc
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Sep4-06, 02:14 AM
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If you happen to have an E&M book handy, there's probably some derivation of the integral form of Gauss's law from the generalized coulomb's law, and then the derivation of the differential form from the integral form, which is fairly trivial using the divergence theorem.

There's a fairly rigorous derivation in Classical Electrodynamics, by Jackson p27-29, and a somewhat less-than-rigorous (but quite descriptive) derivation in Introduction to Electrodynamics, by Grffiths, p65-70
Meir Achuz
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Sep4-06, 08:53 AM
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Quote Quote by Swapnil
I was wondering, how would you prove Gauss's law (either mathematically or intuitively)? I mean, I know that sometimes people take it as the fundamental law (one without proof) but how would you derive such a law from Coulomb's law?

Any help would be extremely appreciated. I have been fussing over this topic for a month now.
Try page 12 of "Classical Electromagnetism" by Franklin for a rigorous, intuitive, mathematical proof.
Swapnil
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Sep5-06, 01:13 PM
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Quote Quote by jbusc
There's a fairly rigorous derivation in Classical Electrodynamics, by Jackson p27-29, and a somewhat less-than-rigorous (but quite descriptive) derivation in Introduction to Electrodynamics, by Grffiths, p65-70
Checked out Jackson. I agree that he gives a fairly rigorous derivation of Gauss' law.
debkar
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Aug9-10, 08:35 AM
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Hey i have a new proof of gauss law.now i am listing it below

let there be an arbitaraly shaped body.let there be a charge Q,e='epsilon not',E=electric field,dS=small surface element.
let the body be consisting of many small surface elements dS1,dS2......
let the individual surfaces be so small that radius here are constant and be r1,r2......... respectively
therefore surface integral of E.dS=E1.dS1+E2.dS2+.....
since theta =0 therefore E.dS=E*dS

therefore putting values

Q*dS1/(4*pi*e*r12)+Q*dS2/(4*pi*e*r22)+....

d(theta1)=dS1/dr12,d(theta2)=dS2/dr22,etc (from definition of solid angle)

taking everything in common


net elctric flux=Q/(4*pi*e)(dS1/dr12+dS2/dr22+...)
=Q/(4*pi*e)(d(theta1)+d(theta2)+....)
=Q/(4*pi*e)(4*pi)=Q/(e)
(since sum of all solid angle around the charge is 4*pi)
GRDixon
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Aug9-10, 05:33 PM
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Quote Quote by Swapnil View Post
I was wondering, how would you prove Gauss's law (either mathematically or intuitively)? I mean, I know that sometimes people take it as the fundamental law (one without proof) but how would you derive such a law from Coulomb's law?

Any help would be extremely appreciated. I have been fussing over this topic for a month now.
In general the electric field of a point charge with arbitrary motion does NOT vary as 1/r^2. For a rigorous formula for the E field of a point charge, see Griffiths, "Introduction to Electrodynamics," 2nd Edition, Eq. 9.107. I have considered the case of a relativistically oscillating charge (wA=.99c) inside a variety of shapes ... sphere, ellipsoid, egg-shape ... and have found (using a computer) that Gauss' law was satisfied in each case. I must confess I was surprised! Others have argued that Gauss' law is one of Maxwell's equations, and is therefore completely general. It appears they're right. I'm convinced that the law holds for all charge motions, including relativistic ones, and for all (fixed) shapes in space (both those enclosing the charge and those external to the charge).
gabbagabbahey
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Aug10-10, 09:13 AM
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Quote Quote by GRDixon View Post
In general the electric field of a point charge with arbitrary motion does NOT vary as 1/r^2. For a rigorous formula for the E field of a point charge, see Griffiths, "Introduction to Electrodynamics," 2nd Edition, Eq. 9.107. I have considered the case of a relativistically oscillating charge (wA=.99c) inside a variety of shapes ... sphere, ellipsoid, egg-shape ... and have found (using a computer) that Gauss' law was satisfied in each case. I must confess I was surprised! Others have argued that Gauss' law is one of Maxwell's equations, and is therefore completely general. It appears they're right. I'm convinced that the law holds for all charge motions, including relativistic ones, and for all (fixed) shapes in space (both those enclosing the charge and those external to the charge).
Not sure why you're reviving a 4 year old thread, but I suppose if you are looking for the Electrodynamic (as opposed to electrostatic) analog to Griffiths' derivation, you could simply apply the Divergence theorem to Jefimenko's equation for [itex]\textbf{E}[/itex]. However, unlike the static case where both Gauss' Law and Coulomb's are easy enough to derive from experiments (empirically), and are hence both valid axioms for the theory, I've never seen Jefimenko's equations derived empirically. Hence, I'd consider Gauss' Law to be more fundamental than Jefimenko's equations (although I suppose you could transform Coulomb's law to a moving reference frame and take Coulomb's law and the two (or 3) axioms of SR as axioms for your theory of electrodynamics), and question the importance of "deriving" Gauss' Law from Jefimenko's equations.
experimentX
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#30
Jun30-12, 05:53 AM
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[itex] \vec E = \frac{\hat r}{4 \pi r^2} \times \frac{Q}{\epsilon_0} \\
\vec E \cdot \hat r 4 \pi r^2 = \frac{Q}{\epsilon_0} \\
\text{Total Flux} = \frac{Q}{\epsilon_0} \\
\oint_s \vec E \cdot \hat n ds = \frac Q{\epsilon_0}[/itex]
Which is integral form of Gauss law.
Killingtensor
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#31
Jun30-12, 09:26 AM
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I'm fairly sure this can be done with lagrangians as long as you are comfortable with the jump from lagrangians to lagrangian densities.


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